Note that \(b \in {\mathcal {C}}^\gamma (\dot{\mathbb {R}}; {\mathcal {L}}(Y,X))\)  and, since the inversion map is analytic, \(a\in {\mathcal {C}}^\gamma (\dot{\mathbb {R}}; {\mathcal {L}}(X,Y))\) .
Before we proceed by induction, we need to consider the cases when \(\gamma = 1,2,3\)  separately.
In the case when \(\gamma =1\) , since \(b(t)a(t) = I\) , by Leibniz' rule, for every \(t\in \dot{\mathbb {R}}\)  we get
\(-a^{\prime }(t) = a(t) b^{\prime }(t) a(t) \qquad (\textrm {in } {\mathcal {L}}(X,Y)),\) 
where
\( b^{\prime }(t) = -2t P + iB + \hat{c} ^{\prime }(t). \) 
By our assumptions we get \( b^{\prime }a, (\cdot )b^{\prime }a \in L^\infty ({\mathcal {L}}(X))\) , which shows that \(a \in \widetilde{\mathfrak {M}}_1({\mathcal {L}}(X,Y))\) .
Similarly, when \(\gamma = 2\) , we get
\(-a^{\prime \prime }(t) = a^{\prime }(t) b^{\prime }(t) a(t) + a(t) b^{\prime \prime }(t) a(t) + a(t) b^{\prime }(t) a^{\prime }(t),\) 
where
\(b^{\prime \prime }(t) = - 2P + \hat{c}^{\prime \prime } (t).\) 
Therefore, our assumptions yield directly that \(b^{\prime \prime }a, (\cdot )^2 b^{\prime \prime } a \in L^\infty ({\mathcal {L}}(X))\)  and, by the step for \(\gamma = 1\) , \(b^{\prime }a^{\prime } \in L^\infty ({\mathcal {L}}(X))\) . Consequently,
\(a \in \widetilde{\mathfrak {M}}_2({\mathcal {L}}(X,Y))\) .
For \(\gamma =3\) , note that
\(-a^{\prime \prime \prime }(t) = & a^{\prime \prime }(t) b^{\prime }(t) a(t) + a^{\prime }(t) b^{\prime \prime }(t) a(t) + a^{\prime }(t) b^{\prime }(t) a^{\prime }(t)\\\nonumber & + a^{\prime }(t) b^{\prime \prime }(t) a(t) + a(t) b^{\prime \prime \prime }(t) a(t) + a(t) b^{\prime \prime }(t) a^{\prime }(t) \\\nonumber & + a^{\prime }(t) b^{\prime }(t) a^{\prime }(t) + a(t) b^{\prime \prime }(t) a(t) + a(t) b^{\prime }(t) a^{\prime \prime }(t),\) 

where
\(b^{\prime \prime \prime }(t) = \hat{c} ^{\prime \prime \prime }(t).\) 
Therefore, the steps for \(\gamma =1\)  and \(\gamma =2\)  show that \(a \in \widetilde{\mathfrak {M}}_3({\mathcal {L}}(X,Y))\) .
Now we can proceed by induction for \(\gamma > 3\) . Fix \(3\le \alpha \le \gamma \) . Then
\(a^{(\alpha )} = \sum _{0 \le k,l,m \le \alpha ,\, l\ne 0 \atop k+l+m = \alpha }a^{(k)}b^{(l)}a^{(m)} \qquad (\textrm {on } \dot{\mathbb {R}}).\) 
By the induction step, for every \(k=0, ..., \alpha - 1\) , the function \((\cdot )^k a^{(k)}\)  is in \(L^\infty ({\mathcal {L}}(X,Y))\) . Moreover, in the case when \(k\ne 0\) , since \(l+m\le \alpha -1\) , by the induction, we have that the functions \((\cdot )^{l+m}b^{(l)}a^{(m)}\in L^\infty ({\mathcal {L}}(X))\) .
Therefore, consider the case of \(k=0\) , that is, \(l+m = \alpha \) . If \(l=\alpha \) , then
\(b^{(\alpha )}=\hat{c}^{(\alpha )}\) , thus \((\cdot )^{\alpha } b^{(\alpha )}a \in L^\infty ({\mathcal {L}}(X))\) .
In the case \(l < \alpha \) , i.e., \(1\le m \le \alpha -1\) , we have
\(b^{(l)} a^{(m)} = \sum _{0 \le m_1,m_2,m_3 \le m, \, m_2\ne 0 \atop m_1+m_2+m_3= m }b^{(l)} a^{(m_1)}b^{(m_2)}a^{(m_3)} \qquad (\textrm {on } \dot{\mathbb {R}}).\) 
Since \(l+m_1\le l+m-m_2\le \alpha -1\)  and \(m_2+m_3\le m \le \alpha -1\) , by the induction step, we get \((\cdot )^{l+m}b^{(l)} a^{(m)} \in L^\infty ({\mathcal {L}}(X))\) . It shows that \(a\in \widetilde{\mathfrak {M}}_\gamma (X,Y)\) .
Applying the formulas obtained for the function \(a\) , one can easily get desired statement for \(a_0\)  and \(a_1\) . We omit details.
The proof in the case of the \({{M}}_\gamma \) -condition follows the one given above.
For the last statement, recall that directly from the definition of \(R\) -boundedness, for any Banach spaces \(X,Y, Z\) , if \(\tau , \sigma \subset {\mathcal {L}}(X,Y)\)  and \(\rho \subset {\mathcal {L}}(Y,Z)\)  are \(R\) -bounded, then the families \(\tau + \sigma \)  and \(\tau \circ \rho \)  are \(R\) -bounded (see e.g. [1]}). Therefore,
the proof of the last statement mimics exactly the argument given already for the usual, norm boundedness.
